It does not gets more complicated than this

It is suspected that $m=7$ so that the $6^{th}$ term will be as follows with the square and fifth roots removed:

$\begin{pmatrix} 7 \\ 5 \end{pmatrix} (\sqrt{2^{\log{(10-3^x)}}})^2 (\sqrt [5] { 2^ { (x-2)\log {3} }})^5 = 21 (2^{\log{(10-3^x)}}) (2^ { (x-2)\log {3} })$

It is in fact, $m = 7$ , because the $2^{nd}$ , $3^{rd}$ and $4^{th}$ coefficients are $7$ , $21$ and $35$ , which are the $1^{st}$ , $3^{rd}$ and $5^{th}$ of an $A.P.$ with a common difference of $7$ .

Now, we have, the $6^{th}$ term equals to $21$ :

$\Rightarrow 21 (2^{\log{(10-3^x)}}) (2^ { (x-2)\log {3} }) = 21 \quad \Rightarrow (2^{\log{(10-3^x)}}) (2^ { (x-2)\log {3} }) = 1$

$\Rightarrow \log{(10-3^x)} + (x-2)\log {3} = 0 \quad \Rightarrow \log{[3^{(x-2)}(10-3^x)]} = 0$

$\Rightarrow 3^{(x-2)}(10-3^x) = 1 \quad \Rightarrow 3^x(10-3^x) = 9 \quad \Rightarrow 10(3^x) -3^{2x} = 9$

$\Rightarrow 3^{2x} - 10(3^x) + 9 = 0 \quad \Rightarrow (3^x - 1)(3^x - 9) = 0$

$\Rightarrow 3^x = 1$ or $9\quad \Rightarrow x = 0$ or $2$

Therefore, the required answer $a+b=0+2=\boxed {2}$

Same solution as Chew seong cheong... Can anyone explain to me how and why reward points change for a question after every attempt.?

a =2, b = 0 sum = 0 + 2 = 2